## Class Room |

When presented with a complex division problem, it is often possible to simplify the problem by factorizing the divisor. We saw this in the previous exercise, division by an even number, where we saw that a complex division problem could be made simpler by dividing both the divisor and dividend by 2, 4 or 8.

In this exercise, we will be doing the same thing, but with different factors, like 7 or 9. In order to do that, we must be able to quickly tell if the divisor and the dividend are divisible by the factor. The below table lists out the techniqies that can be used.

Divisible by | When | Example |
---|---|---|

2 | Last digit is even (0,2,4,6,8) | 12,24 1992 are all divisible by 2 because the last digit is even |

3 | If the sum of the digits is divisible by 3 | Both 12 (1+2=3) and 192 (because 1 + 9 + 2 = 12) are divisible by 3 |

4 | The last 2 digits are divisible by 4 | t22 |

5 | The last digit is 0 or 5 | t22 |

6 | The last 2 digits are divisible by 4 | t22 |

7 | The last 2 digits are divisible by 7 | t22 |

8 | The last three digits are divisible by 8 | 6112 is divisible by 8 because the last 3 digits, 112÷8=14, is divisible by 8. So is 1008 because 008 is divisible by 8. |

9 | The sum of all the digits is divisible by 9 | 6273 is divisible by 9 because 6+2+7+3=18, which is divisible by 9. |

10 | The number must end in 0 | 1230 is divisible by 10 because it ends with a 0. So is 137000. |

11 | Sum all the odd number digits and then subtract the sum of the even number digits and the result is either 0 or divisible by 11 | Check 1353. Sum of odd digits is (3+3=6) and the sum of even digits is (1+5=6). Now, 6-6=0, so 1353 is divisible by 11. Check 1749. Sum of odd digits (7+9=16) and sum of even digits (1+4=5). Now, 16-5=11, so 1749 is divisible by 11. |

12 | The number is divisible by both 3 and 4 | 72 is divisible by 12 because it is divisible by 3 (72 ÷ 3 = 24) and it is also divisible by 4 (72 ÷ 4 = 18) |

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